Questions: Chain Homotopy and Chain Equivalence

Question 1

If f and g are chain homotopic chain maps (f ~ g via chain homotopy P), why do they induce the same map on homology?

• A. Because P is an isomorphism
• B. Because for any cycle z (with dz = 0), f(z) - g(z) = dP(z) + Pd(z) = dP(z) + 0 = dP(z), which is a boundary — so [f(z)] = [g(z)] in homology
• C. Because chain homotopic maps are equal
• D. Because the chain groups are free abelian

Answer: (B)

The chain homotopy equation f - g = dP + Pd evaluated on a cycle z (with dz = 0) gives f(z) - g(z) = dP(z) + P(dz) = dP(z) + P(0) = dP(z). So f(z) and g(z) differ by a boundary dP(z), meaning they represent the same homology class. This is the chain-level version of 'homotopic maps induce the same map on homology.' The argument works for any chain homotopy, regardless of whether it comes from a topological homotopy.

Question 2

Chain homotopy is an equivalence relation on chain maps from C_* to D_*.

• T. True
• F. False

Answer: True

Reflexive: f ~ f via P = 0 (since f - f = 0 = d·0 + 0·d). Symmetric: if f ~ g via P (f - g = dP + Pd), then g - f = -(dP + Pd) = d(-P) + (-P)d, so g ~ f via -P. Transitive: if f ~ g via P and g ~ h via Q, then f - h = (f - g) + (g - h) = dP + Pd + dQ + Qd = d(P + Q) + (P + Q)d, so f ~ h via P + Q. These are straightforward algebraic verifications.

Question 3

A chain map f: C_* → D_* is a chain homotopy equivalence if there exists a chain map g: D_* → C_* with g ∘ f ~ id_{C_*} and f ∘ g ~ id_{D_*}. Such maps always induce isomorphisms on homology.

• T. True
• F. False

Answer: True

If g ∘ f ~ id, then (g ∘ f)_* = g_* ∘ f_* = id_* on homology. Similarly f_* ∘ g_* = id_*. So f_* and g_* are inverse isomorphisms. This means chain homotopy equivalent complexes have isomorphic homology. The converse is not generally true: two chain complexes with isomorphic homology need not be chain homotopy equivalent (just as two spaces with the same homology need not be homotopy equivalent).

Question 4

Explain how the chain homotopy for the proof that homotopic maps induce the same map on homology is constructed from a topological homotopy H: X × [0,1] → Y.

Model answer: The prism operator P_n: C_n(X) → C_{n+1}(Y) is defined using a triangulation of Δ^n × [0,1]. Each singular n-simplex σ: Δ^n → X is 'thickened' to a prism σ × id: Δ^n × [0,1] → X × [0,1], and composing with H gives a map to Y. Triangulating the prism Δ^n × [0,1] into (n+1)-simplices gives a chain in C_{n+1}(Y). The boundary of this prism chain, by geometry, equals f#(σ) - g#(σ) plus lower-dimensional prism terms, yielding the chain homotopy equation f# - g# = dP + Pd.

The prism operator is the algebraic distillation of the geometric fact that a homotopy sweeps out a prism. The triangulation of Δ^n × [0,1] into (n+1)-simplices is standard (using n+1 simplices), and the boundary formula for this triangulation directly gives the chain homotopy relation. This construction is the technical heart of the proof of homotopy invariance of singular homology.